Class 12 Amines — Preparation, Basicity and Reactions
Amines are the last big functional-group chapter of Class 12, and they reward students who understand one idea: the nitrogen atom has a lone pair, and everything depends on how available that lone pair is. Make it more available and the amine is a stronger base and a better nucleophile. Tie it up — in a benzene ring, or with an acetyl group — and the compound calms down. Preparation routes, basicity order, the identification tests and the whole of diazonium chemistry all follow from that single sentence.
Classification and the group itself
Base behaviour: R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH⁻, Kb = [RNH₃⁺][OH⁻] ÷ [RNH₂]
- Amines are classified by how many carbon groups are on nitrogen — not by the carbon skeleton. Note the contrast with alcohols, where the classification depends on the carbon. tert-Butylamine, (CH₃)₃C–NH₂, is a primary amine because only one carbon touches the nitrogen.
- Kb is the base dissociation constant; pKb = −log Kb. Smaller pKb means a stronger base.
- 1° and 2° amines have N–H bonds and hydrogen bond to each other; 3° amines have none, so they boil at lower temperatures than isomeric 1° and 2° amines despite the same molar mass.
Preparation — six routes and what each is for
| Route | Reagents | Gives | Carbon count |
|---|---|---|---|
| Reduction of nitro compounds | H₂/Pd, or Sn + HCl, or Fe + HCl | 1° amine (the standard route to aniline) | Same |
| Ammonolysis of alkyl halides | NH₃ in a sealed tube | Mixture of 1°, 2°, 3° and the quaternary salt | Same |
| Reduction of nitriles | LiAlH₄, or H₂/Ni | 1° amine | One more |
| Reduction of amides | LiAlH₄ | 1° amine | Same |
| Gabriel phthalimide synthesis | Potassium phthalimide + R–X, then hydrolysis | Pure 1° amine only | Same |
| Hoffmann bromamide degradation | Amide + Br₂ + 4 NaOH | 1° amine | One less |
The last column is the exam-critical one. Two of these routes change the number of carbon atoms in opposite directions, and questions exploit that constantly.
Count it: left has 2 C, 5 + 4 = 9 H, 1 N, 1 + 4 = 5 O, 2 Br, 4 Na. Right has 1 + 1 = 2 C, 5 + 4 = 9 H, 1 N, 3 + 2 = 5 O, 2 Br, 2 + 2 = 4 Na. Balanced — and the amine has one carbon fewer than the amide, because that carbon leaves as carbonate.
Two limitations worth stating in an answer:
- Ammonolysis gives a mixture, because each amine formed is itself a nucleophile and attacks more halide. Only a large excess of ammonia favours the primary amine, and even then separation is needed.
- Gabriel synthesis fails for aromatic amines. It needs the phthalimide anion to displace a halide, and aryl halides do not undergo that substitution under these conditions. So aniline cannot be made this way — the nitrobenzene reduction route is used instead.
Basicity — the order is not what you first expect
On paper, alkyl groups push electron density onto nitrogen (+I), so more alkyl groups should mean a stronger base: 3° > 2° > 1° > NH₃. In the gas phase that is exactly what is observed. In water it is not, because two more factors join in:
- Solvation. The protonated ion R₃NH⁺ has only one N–H to hydrogen bond with water, while RNH₃⁺ has three. Better hydration means a more stable cation, which means a stronger base.
- Steric crowding. Three bulky groups make it physically harder for a proton and for water molecules to reach the nitrogen.
Push (+I) helps, crowding and poor solvation hurt, and the compromise gives these orders in aqueous solution:
Ethyl series: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃
Both orders put the secondary amine first — that is the reliable takeaway. Aromatic amines sit far below all of these. In aniline the nitrogen lone pair is delocalised into the benzene ring, so it is simply not available to accept a proton. Adding electron-withdrawing groups (–NO₂) weakens aniline further; electron-donating groups (–CH₃, –OCH₃) strengthen it slightly.
Worked example 1 — how basic is an amine, in numbers?
Question. Calculate the pH of 0.10 M methylamine (Kb = 4.4 × 10⁻⁴) and of 0.10 M aniline (Kb ≈ 4.3 × 10⁻¹⁰). Take Kw = 1.0 × 10⁻¹⁴ at 25 °C.
Methylamine, quick route.
[OH⁻] = √(Kb × C) = √(4.4 × 10⁻⁴ × 0.10) = √(4.4 × 10⁻⁵) = 6.63 × 10⁻³ mol/L
pOH = −log(6.63 × 10⁻³) = 3 − 0.822 = 2.18
pH = 14.00 − 2.18 = 11.82
Cross-check, because α is not small here.
α = 6.63 × 10⁻³ ÷ 0.10 = 6.6%, above the usual 5% comfort line, so solve properly:
x² + (4.4 × 10⁻⁴)x − 4.4 × 10⁻⁵ = 0
Discriminant = 1.936 × 10⁻⁷ + 1.76 × 10⁻⁴ = 1.76194 × 10⁻⁴, √ = 1.32738 × 10⁻²
x = (−4.4 × 10⁻⁴ + 1.32738 × 10⁻²) ÷ 2 = 6.42 × 10⁻³ mol/L
pOH = 2.19, pH = 11.81
The two routes agree to within 0.01 pH unit, so the approximation was acceptable — but you only know that after checking.
Aniline.
[OH⁻] = √(4.3 × 10⁻¹⁰ × 0.10) = √(4.3 × 10⁻¹¹) = 6.56 × 10⁻⁶ mol/L
pOH = 6 − 0.817 = 5.18 → pH = 14.00 − 5.18 = 8.82
Read the result. The same concentration gives pH 11.8 for methylamine and only 8.8 for aniline — three whole pH units, a factor of about a thousand in hydroxide concentration. Resonance delocalisation of one lone pair does that. This is the number to quote when a question asks you to justify "aniline is a much weaker base than methylamine".
The identification tests
| Test | Primary amine | Secondary amine | Tertiary amine |
|---|---|---|---|
| Carbylamine (CHCl₃ + alcoholic KOH) | Foul-smelling isocyanide | No reaction | No reaction |
| Hinsberg (C₆H₅SO₂Cl, then NaOH) | Product dissolves in alkali | Product does not dissolve in alkali | No reaction at all |
| HNO₂ (NaNO₂ + HCl, cold), aliphatic | Brisk N₂ gas; alcohol formed | Yellow oily nitrosamine | Forms a soluble salt |
| HNO₂, aromatic at 273–278 K | Stable diazonium salt | Yellow oily nitrosamine | Salt formation |
The Hinsberg result has a clean explanation worth writing out: a primary amine gives a sulphonamide that still has one N–H, and that N–H is acidic enough (the sulphonyl group withdraws strongly) to be removed by NaOH, giving a soluble salt. A secondary amine's product has no N–H left, so nothing dissolves. A tertiary amine has no N–H to react with in the first place.
Count: left 2 C, 5 + 1 + 3 = 9 H, 1 N, 3 Cl, 3 K, 3 O. Right 2 C, 3 + 6 = 9 H, 1 N, 3 Cl, 3 K, 3 O. Balanced.
Safety note: isocyanides are extremely foul-smelling and toxic. The carbylamine test is a fume-cupboard demonstration, not a student bench experiment.
Diazonium salts — one intermediate, many products
Aniline plus nitrous acid at 273–278 K gives benzenediazonium chloride. The low temperature is not optional: above about 278 K the salt decomposes to phenol and nitrogen gas.
Count: left 6 C, 7 + 2 = 9 H, 2 N, 2 O, 1 Na, 2 Cl. Right 6 C, 5 + 4 = 9 H, 2 N, 2 O, 1 Na, 2 Cl. Balanced.
This single salt is the gateway to substituents you cannot install by direct substitution:
| Reagent on C₆H₅N₂⁺Cl⁻ | Product | Name of reaction |
|---|---|---|
| CuCl / HCl or CuBr / HBr | Chlorobenzene or bromobenzene | Sandmeyer |
| Copper powder + HX | Halobenzene | Gattermann |
| CuCN / KCN | Benzonitrile | Sandmeyer |
| KI | Iodobenzene | — (no copper needed) |
| HBF₄, then heat | Fluorobenzene | Balz–Schiemann |
| H₃PO₂ + water, or ethanol | Benzene (the group is simply removed) | — |
| Warm water | Phenol | — |
| Phenol in mild alkali | Orange azo dye | Coupling |
| Aniline in mild acid | Yellow azo dye | Coupling |
Coupling reactions are electrophilic substitutions on the other ring, and they need an activated ring (phenol or an amine). They occur at the para position when it is free.
Worked example 2 — yield in a Hoffmann degradation
Question. 5.91 g of ethanamide (acetamide) is treated with bromine and sodium hydroxide. 2.52 g of methylamine is collected. Find the percentage yield.
Step 1 — molar masses (C 12.011, H 1.008, N 14.007, O 15.999):
M(CH₃CONH₂) = 24.022 + 5.040 + 14.007 + 15.999 = 59.07 g/mol
M(CH₃NH₂) = 12.011 + 5.040 + 14.007 = 31.06 g/mol
Step 2 — moles. The equation is 1 : 1, so
n = 5.91 ÷ 59.07 = 0.100 mol of amide → 0.100 mol of amine
Step 3 — theoretical mass.
0.100 × 31.06 = 3.11 g
Step 4 — percentage yield.
(2.52 ÷ 3.11) × 100 = 81.1%
Sanity check on the chemistry, not just the arithmetic: the product mass must be smaller than the starting mass here, because a whole carbon and an oxygen leave as carbonate. If your "theoretical yield" came out heavier than the amide, you have used the wrong molar mass.
Worked example 3 — a three-step conversion
Question. Convert aniline into 4-bromoaniline without getting the tribromo product.
The problem. Aniline with bromine water gives 2,4,6-tribromoaniline directly, because –NH₂ activates the ring so strongly that all three free positions react. You cannot stop at one bromine.
Step 1 — protect. Acetylate the amine with acetic anhydride to give acetanilide. The nitrogen lone pair is now shared with the acetyl carbonyl, so the ring is activated much less.
Step 2 — brominate. Br₂ in ethanoic acid now gives mainly p-bromoacetanilide, a single substitution.
Step 3 — deprotect. Hydrolyse with acid or alkali to remove the acetyl group, giving 4-bromoaniline.
The same protect-react-deprotect logic explains why nitration of aniline is done on acetanilide: in strong nitrating acid, aniline is largely protonated to the anilinium ion, which is deactivating and meta directing, so the product mixture contains an appreciable amount of the meta isomer. Protecting the nitrogen avoids that.
Common mistakes that cost marks
- Quoting 3° > 2° > 1° as the basicity order in water. The secondary amine is the strongest of the three in aqueous solution. Mention solvation and steric hindrance, or the mark is not given.
- Classifying amines by the carbon. tert-Butylamine is a primary amine. Count the groups on nitrogen.
- Losing track of carbon count. Hoffmann bromamide loses one carbon; nitrile reduction gains one. Questions are built around this.
- Using Gabriel synthesis for aniline. It does not work for aromatic amines.
- Attempting Friedel–Crafts on aniline. The Lewis acid catalyst bonds to the nitrogen lone pair, deactivating the ring — the reaction fails.
- Forgetting the temperature for diazotisation. Without "273–278 K" the answer is incomplete, because the salt decomposes above that range.
- Saying the carbylamine test detects all amines. Only primary amines, aliphatic and aromatic, respond.
- Treating a nitrosamine as a harmless product. Many N-nitroso compounds are known carcinogens; the test is written about far more often than it is performed.
Why this chapter matters
| Where it appears | What is actually tested |
|---|---|
| CBSE / ICSE Class 12 organic unit | Basicity order with reasons, distinguishing 1°/2°/3°, conversions |
| Multi-step conversions | Aniline → diazonium → almost any substituted benzene |
| Practical and viva | Nitrogen detection, carbylamine and Hinsberg reasoning |
| NEET / JEE foundation | Ranking basicity, predicting diazonium products |
| IIT-JAM / CUET-PG later on | Mechanisms, protecting groups, azo-dye chemistry |
Check your board's current syllabus document for your own session rather than relying on an older question pattern.
Turn "stronger base" into a number. Feed a Kb and a concentration into the pH / pOH calculator and watch methylamine land near pH 11.8 while aniline barely reaches 8.8. Seeing the gap makes the resonance explanation stick far better than reading the order once more.
Open the pH / pOH Calculator →Organic conversions are where most Class 12 students lose marks, and they are fixable with structured practice. ABC Chemistry teaches Class 11–12 chemistry at the Gurugram centre and online for students across India — details at abcchemistry.in.